The radius of the semicircle having an arc of length equal to 8.5 [tex]\pi[/tex] in a right-angle triangle ABC is equal to 7.5 cm.
Given:
Angle ABC of triangle ABC is a right angle. The sides of ABC are the diameters of semicircles.
The area of the semicircle on AB equals 8[tex]\pi[/tex].
Area of a semicircle = [tex]\pi r^2/2[/tex]
Therefore:
[tex]\pi r^2/2[/tex] = 8[tex]\pi[/tex]
[tex]r^2 = 16[/tex]
[tex]r = 4[/tex]
Next, the arc of the semicircle on AC has a length of 8.5[tex]\pi[/tex].
Length of the arc of a semicircle = [tex]\pi r[/tex]
[tex]\pi r[/tex] = 8.5[tex]\pi[/tex]
[tex]r = 8.5[/tex]
Using Pythagoras theorem
[tex]8.5^{2} = 4^{2} + x^{2}[/tex]
[tex]x^{2} = 8.5^2 - 4^2\\[/tex]
[tex]x^{2} = 56.25[/tex]
[tex]x = \sqrt{56.25}[/tex]
[tex]x = 7.5[/tex]
The radius of the semicircle of BC = 7.5 Units.
Refer to this complete question for this:
Angle ABC of triangle ABC is a Right angle. The sides of ABC are the diameters of semicircles as shown. The area of the semicircle on AB equals 8pi and the arc of the semicircle on AC has a length of 8.5pi. What is the radius of the semicircle of BC?
To learn more about the Pythagoras theorem visit: https://brainly.com/question/343682
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